Let $G$ be an infinite, virtually cyclic group, i.e., $G$ has an infinite cyclic subgroup $H$ of finite index.
I have a question: Why is there a finite bound on the orders of finite subgroups of $G$?
What I tried: If $G$ is finite, then $|G|$ is a bound on the orders of its subgroups. But I don't know why is true about an infinite virtually cyclic group.
Best Answer
The first step is to realize that $G$ not only has a finite index infinite cyclic subgroup $H$, but in addition it has a finite index infinite cyclic normal subgroup $N$, namely $N = \bigcap_{g \in G} g H g^{-1}$: it is a general group theory exercise that for any group $G$ and any finite index subgroup $H < G$, the subgroup $N$ as defined is finite index and normal; and since $N < H$ it follows that $N$ is cyclic.
So suppose now that $C \subset G$ is a subgroup whose order is larger than the order of the finite quotient group $G/N$. The kernel of the homomorphism $C \mapsto G/N$ is therefore nonempty. But the kernel of that homomorphism is $C \cap N$, and so any nontrivial element of the kernel has infinite order, implying that the subgroup $C$ is infinite.